The aldol condensation of $CH_{3}CHO$ results in the formation of
- A$CH_{3}CH(OH)CH_{2}CHO$
- B$CH_{3}CH_{2}CH(OH)CHO$
- C$CH_{3}CH_{2}CH_{2}OH$
- D$CH_{3}CH(OH)CH_{3}$
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Similar Questions
The major product of the following reaction sequence is:
Reactant $A$ of the above reaction is
$(A)$ + $\begin{matrix} CH_2-OH \\ | \\ CH_2-OH \end{matrix}$ $\xrightarrow{\text{Pyridine}} \begin{matrix} CH_2-O \\ | \\ CH_2-O \end{matrix} > C = O$
$(A)$ + $\begin{matrix} CH_2-OH \\ | \\ CH_2-OH \end{matrix}$ $\xrightarrow{\text{Pyridine}} \begin{matrix} CH_2-O \\ | \\ CH_2-O \end{matrix} > C = O$
Medium
View SolutionMatch the following lists:
The major product $'P'$ formed in this reaction is:
Medium
View SolutionMatch the following reactions in List-$I$ with their corresponding final products in List-$II$:
List-$I$ (Type of reaction) List-$II$ (Final Product) $A$. Reimer-Tiemann reaction $I$. Chlorobenzene $B$. Etard reaction $II$. Salicylaldehyde $C$. Sandmeyer reaction $III$. Benzaldehyde $D$. Friedel-Crafts reaction $IV$. Acetophenone
The correct answer is:
| List-$I$ (Type of reaction) | List-$II$ (Final Product) |
|---|---|
| $A$. Reimer-Tiemann reaction | $I$. Chlorobenzene |
| $B$. Etard reaction | $II$. Salicylaldehyde |
| $C$. Sandmeyer reaction | $III$. Benzaldehyde |
| $D$. Friedel-Crafts reaction | $IV$. Acetophenone |
The correct answer is:
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